1. Different gametes are only going to be produced from the allelic pairs that are
different. Since alignment during metaphase I of meiosis should be random for
unlinked gene loci, each allele is equally probable of ending up in the same cell.
A
a
B
b
B
One heterozygous locus- two different gametes; two
heterozygous loci- 4 different gametes; three
heterozygous loci- 8 different gametes, etc.
For 7 heterozgyous loci- 27 different gametes, or 128.
Ans: (c).
b
2. Probability has no memory. The likelihood of having two girls will follow the
product rule for all unborn children. The probability of a girl is ½. So the probability
of two girls in two births would be ½ x ½ = ¼.
Ans: (d)
3. The two true-breeding parents would produce a hybrid (heterozygote).
In a cross to a homozygous recessive parent (backcross) only the heterozygous parent
will be able to pass on dominant alleles.
AaBb x aabb
The heterozgous parent will produce 4 types of gametes in equal frequency:
¼ AB, ¼ Ab, ¼ aB, ¼ ab
Fertilization with the AB gamete will lead to heterozygotes in the next generation.
Ans: ¼ (c)
4. To first answer this question, look to see if the recessive trait is observed in the
progeny- if so, then the parents must carry a recessive allele. Both wild-type and
mutant phenotypes are observed in the progeny, so at least one of the parents must be
heterozygous.
Next observe the frequencies that the dominant and recessive traits are inherited. If
one of the parents is homozygous recessive for a trait, we would expect the trait to be
inherited in 50% of the offspring. If both parents are heterozygous for the trait, we
would expect a 3:1 ratio of dominants to recessives
For the eye trait, 4/8 (i.e. ½) are wild type, and 4/8 are sepia.
For the wing trait, 6/8 (i.e. ¾) are wild type, and 2/8 are vestigial
Therefore, the parents would be expected to be:
se+/se vg+/vg x se/se vg+/vg
Ans: (e)
5. See page 3 in Hartwell and figure 1.12
Ans: (a)
AaBbCcDdEe x AaBbccDdEE
6. P(A_B_C_D_E_) =
¾ x ¾ x ½ x ¾ x 1 = 27/128 Ans: (c)
7. P(A_B_C_D_E_ or A_B_ccD_E_) =
27/128 + 27/128 = 54/128 = 27/64 Ans: (c)
8. P(neither parent) = 1- P(either parent)
1-27/64 = 37/64 Ans: (d)
9. Probability of affected child in III-2 x III-6 marriage:
aa
AA
Aa
aa
AA
½ Aa
P (affected child) = P(III-6) is a heterozygote x P(III-6 will pass recessive allele to a
gamete)
P ( ½ x ½ ) = ¼ Ans: (b)
10. Probability of affected child in III-1 x III-7 marriage:
AA
A
a
A AA Aa
a Aa aa
aa
Aa
2/3Aa
aa
Aa
AA
Aa
AA
½ Aa
P (affected child) = P(III-1) is a heterozygote x P(III-7 is a heterozygote) x
P(heterozygous parents will both pass on recessive alleles to offspring)
P ( 2/3 x ½ x ¼) = 1/12 Ans: (e)